Question

Limits....again. Help!

Answer 1

Both numbers 6 and 7, but let's begin with 6!!

Answer 2

@peachpi @zepdrix :)

Answer 3

L'hospital for first one

Answer 4

#6) b = 4?

Answer 5

Ya :)
Imma wait and see what Ganeshie has to say though ^^

Answer 6

Notice that the denominator is 0 when you plugin x=0
Also, for the limit to exist (finite), you must get the indeterminate 0/0 form when you plugin x=0.
otherwise the limit does not exist (infinite).

so, the numerator part must evaluate to 0 at x=0

Answer 7

\(\sqrt{ax+b}-2\)

evaluating it at \(x=0\) and setting equal to \(0\) does give you \(b=4\)

Answer 8

plugin the \(b\) value, rationalize and you will see what \(a\) needs to be

Answer 9

For 7 note this you can do the same with |2x-1| you'll have an ugly numerator at the beginning, but it'll simplify to something nice :D